Applications of Exponential and Log Equations
Exponential and Logarithmic functions have perhaps more real-world applications than
any other class of functions at the pre-calculus level and beyond. These functions
govern population increase as well as interest income in a bank. And since (it seems)
virtually everything decays exponentially, we can apply exponential decay equations to
many different applications (see http://www.mathmotivation.com/science/carvalue.html )
Exponential Growth and Decline:
Exponential growth and decline is governed by the function: N(t) = Noekt
where
No = the original amount at time t=0.
k = the (decimal) rate of growth per time period, be it years, days, etc.
t = the units used in the time period.
N(t) = the amount at time t
NOTE: N(t) means “N as a function of t”, NOT “N times t”!
Example: A population grows exponentially 3% per year. If the population is initially
1000, how many years will it take for the population to double to 2000? How many years
will it take for the population to reach 4000?
Since the original amount is 1000, No=1000. Also, we know that N(t) = 2000 and also
N(t) = 4000. In both cases, k = 0.03 . So we have to solve the equations:
2000 = 1000e0.03t and
0.03t
4000 = 1000e
.
Here are the solutions:
2000 = 1000e0.03t
2 = e0.03t
LN(2) = LN(e0.03t)
LN(2) = 0.03t  LN(e)
LN(2) = 0.03t
[LN(2)]/0.03 = t  23.1 yrs
Given
Division Property of Equality
If A = B, LN(A) = LN(B)
Power Rule For Logs
Loga(a) = 1
Division Property of Equality
4000 = 1000e0.03t
4 = e0.03t
LN(4) = LN(e0.03t)
LN(4) = 0.03t  LN(e)
LN(4) = 0.03t
[LN(4)]/0.03 = t  46.2 yrs
Given
Division Property of Equality
If A = B, LN(A) = LN(B)
Power Rule For Logs
Loga(a) = 1
Division Property of Equality
So it took a little over 23 years for the population to double. The population doubled
again after another 23 years. As mentioned in the previous section, exponential growth
results in a doubling at a fixed interval of time.
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Example: A car’s value is declining exponentially. The car is currently 3 years old and
has a value of $18000. The car sold for $26,000 brand new. How much will the car be
worth 5 years from now?
Since the new price was $26,000, No = 26000. This gives us
N(t) = 26000ekt
We can use the given data to solve for k by noting that at t=3, N(3) = 18000. This gives
us
18000 = 26000ek3
Now, solve for k.
18000/26000 = e3k
18/26 = e3k
LN(18/26) = LN(e3k)
LN(18/26) = 3kLN(e)
LN(18/26) = 3k
[LN(18/26)]/3 = k  -0.1226
Division Property of Equality
Simplify Fraction
If A = B, LN(A) = LN(B)
Power Rule For Logs
Loga(a) = 1
Division Property of Equality
So now, the completed model is N(t) = 26000e-0.1226t
We can now use the completed model to answer our question, “How old will the car be
in 5 years?” Since the car is already 3 years old, in 5 years it will be 8 years old, so we
let t = 8 in our completed model to get
N(8) = 26000e-0.12268
= $9750, rounded
You can verify this answer with the Car Value Calculator, shown at
http://www.mathmotivation.com/science/carvalue.html
You will note that the answer is close, but not an exact match – this is due to the fact
that we rounded our value of k more than the calculator does.
General Procedure For Exponential Growth & Decay Applications
If you noticed, in the last problem, we followed this procedure:
1. Use the given data to solve for the constants k and No, unless these constants
are already given.
2. Write the completed model.
3. Use the completed model to predict other values.
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Compound Interest – Solving For t
Now that we are able to solve for variables that are exponents, we may solve the
compound interest formula A = P(1 + r/n)nt for t. In other words, we can calculate the
time required for a given amount of money P to reach some given future value A.
Example: A person places $4000 in a bank account that earns 5% interest
compounded monthly. How many years will it take for the total value to reach
$100,000?
We let P = 4000, r=0.05, A = 100000, and n = 12 in A = P(1 + r/n)nt to get
100000 = 4000(1 + 0.05/12)12t
Now, solve this.
100000/4000 = (1 + 0.05/12)12t
25 = (1 + 0.05/12)12t
LN(25) = LN[(1 + 0.05/12)12t]
LN(25) = 12tLN(1 + 0.05/12)
LN(25)/LN(1 + 0.05/12) = 12t
[LN(25)/LN(1 + 0.05/12)]/12 = t
t = 64.5 years (rounded)
Division Property of Equality
Simplify Fraction
If A = B, LN(A) = LN(B)
Power Rule For Logs
Division Property of Equality
Division Property of Equality
Evaluate Logs and divide
You may check this answer by plugging back into the original equation
100000 = 4000(1 + 0.05/12)12t
Use of the Log Scales
Base-10 log scales are used in different areas of science. Common applications include
the pH scale and the Richter scale. An example involving pH is included here.
Example: pH is commonly used to define acidity in a chemical solution. pH is defined as
pH = -LOG10[H+]. If pH is 5.2, what is [H+]?
Letting pH = 5.2 results in 5.2 = -LOG10[H+]. We multiply both sides by –1 to get
-5.2 = LOG10[H+]
Now, solve this log equation.
-5.2 = LOG10[H+]
10-5.2 = [H+]
[H+] = 0.000006309
[H+] = 6.309 x 10-6
Given
Rewrite in equivalent exponential form
Evaluate power of 10
Rewrite in scientific notation
Did You Know This About pH? The lower the pH, the more acidic a solution is. Since
pH uses a base-10 logarithmic scale, it takes a 10-fold increase in acidity to lower the pH
1 unit. With respect to the current issue of “acid rain”, if a lake’s pH decreases by 1 unit,
the acidity of a lake increases by a power of 10!
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